The radioactive decay of a substance is given by $R = 1000 e^{-ct}, \, t \geq 0.$ (a) Find the number of atoms when the substance started to decay - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 6

Given that it takes 5730 years for half of the substance to decay, we set up the equation: $1000 e^{-5730c} = 500.$

Dividing both sides by 1000 yields: $e^{-5730c} = \frac{1}{2}.$

Taking the natural logarithm of both sides: $-5730c = \ln{\left(\frac{1}{2}\right)} \\ c = -\frac{\ln{\left(\frac{1}{2}\right)}}{5730}.$

Calculating this gives: $c \approx 0.000121.$

Thus, the value of c to 3 significant figures is 0.000121.

To find the number of atoms left at $t = 22 \, 920$, we substitute into the equation: $R = 1000 e^{-c(22920)}.$

Using the value of c found in part (b): $R = 1000 e^{-0.000121 \times 22920}.$

First, calculate the exponent: $-0.000121 \times 22920 \approx -2.77.$

Now substituting this back, we get: $R \approx 1000 e^{-2.77} \approx 62.5.$

Therefore, the number of atoms left when $t = 22 \, 920$ is approximately 62.5.

The graph of R against t is an exponential decay curve starting at 1000 and approaching 0 as t increases.

• The y-axis represents the number of atoms, R, starting at 1000.
• The x-axis represents time, t.
• The curve should be smooth, continuously decreasing, reflecting the nature of radioactive decay, while never quite reaching zero.

Label the axes appropriately and indicate that at $t = 5730$, $R$ will be at 500.